How can I separate or solve (eg for v) this ecuation: (c+v/c-v)=e???

Question

ganeshie8 · Answer

$\large   \frac{c+v}{c-v} = e$

ganeshie8 · Answer

like that ?

anonymous · Answer

yes

ganeshie8 · Answer

heard of componendo-dividendo before ?

anonymous · Answer

no

ganeshie8 · Answer

its a simple, but very interesting property

ganeshie8 · Answer

here it is :

if  $\large \frac{a}{b} = \frac{c}{d} $ , then :

$\large  \frac{a+b}{a-b} = \frac{c+d}{c-d}$

ganeshie8 · Answer

http://en.wikipedia.org/wiki/Componendo_and_dividendo

ganeshie8 · Answer

$\large   \large   \frac{c+v}{c-v} = e$

$\large   \large   \frac{c+v}{c-v} = \frac{e}{1}$

applying componendo-dividendo :

$\large   \large   \frac{2c}{2v} = \frac{e+1}{e-1}$

$\large   \large   \frac{c}{v} = \frac{e+1}{e-1}$

ganeshie8 · Answer

next, u may separate $v$ by cross multiplying

ganeshie8 · Answer

see if it makes some sense

anonymous · Answer

I believe you and I'm going to verify :) Thanks so much... you made my day.

ganeshie8 · Answer

np :) i felt the same when i first came to knw about "componendo-dividendo" property
it has brighten up my 2-3 days in a row xD

ganeshie8 · Answer

another fav of componendo-divedendo is @ParthKohli 
he may add something to this...

parthkohli · Answer

$$\dfrac{c+v}{c-v}=e$$$$\Rightarrow c+v= ec - ev$$$$\Rightarrow c - ec = -ev - v$$$$\Rightarrow c(e - 1) = v(e + 1)$$$$\Rightarrow v = \dfrac{c(e-1)}{e+1}$$

anonymous · Answer

Thanks a lot, you gave another way of solving... I'd been trying not knowing the "componendo-dividendo" (ganeshi told how), but your way is easier. Thanks again!!!